The equation
a!
b!=
c! is trivial if we allow
a or
b to be 0 or 1.
However, adding that 1<a<b<c, also allows trivial solutions: which? What condition should be added to disallow such solutions?
Finally, adding that condition, can you find any solution to the problem?
(Q1) However, adding that 1<a<b<c, also allows trivial solutions: which?
(A1) Trivial solutions like:
(3!)*(5!) =(6!)
(4!)*(23!) =(24!)
(5!)*(119!) = (120!)
occurs indefinitely.
(Q2) What condition should be added to disallow such solutions?
(A2) The trivial solutions inclusive of (A1) occurs by taking:
{b, c} = {(a-1)!, a!}
so imposing the conditions:
b is NOT equal to (a-1)! and,
c is NOT equal to a!
will certainly disallow the trivial solutions under reference.
(Q3) Finally, adding that condition, can you find any solution to the problem?
(A3) I know of only the solution:
6!*7!=10!
Puzzle Solution
